Probability says it’s only 80% raining today

The following is paraphrased from an equally short passage in The Economist’s “Numbers Guide”, the best little book on numbers available to mankind.

It Certainly Seems to Be Raining

It’s not 80% raining

Wunderground.com says there’s an 80% chance of precipitation today. But it’s raining, so, wunderground is wrong. There is 100% certainty that it is raining.

Probability varies from 0 to 1, where 0 is when something is definitely not going to happen, and 1 means something certainly will happen. The probability that it will rain today is 1.

It’s up in the air

If I flip a coin, there are two possible outcomes: heads or tails. The coin is equally likely to land on either side. So the chance of me getting heads is one out of two possible outcomes, in other words, 1/2.

If I flip the coin twice, the chance of getting two heads is the multiple of the probability of each outcome: 1/2 x 1/2, which is 1/4, or 0.25.

If I flip the coin three times, what’s the probability of getting two (and only two) heads? To solve this problem, first think of how many possible outcomes there are. The three tosses could have any of the following outcomes: HHH, THH, HTH, HHT, TTH, HTT, THT, or TTT. So, as you can see, three of these 8 possible outcomes have two heads: THH, HTH, and HHT. That means my chances of getting two heads with three tosses is 3/8 = 0.375.

In general, the probability of any event, A, is:

P(A) = n_A / n

where n_A is the number of possible outcomes where event A occurs, and n is the total number of possible outcomes.

The total number of possible outcomes, n, is easy to calculate mathematically: in this example, it’s the number of possible outcomes of each toss multiplied by the number of possible outcomes of each other toss. There are two possible outcomes of each toss, and there are three tosses, so the total possible outcomes is 2*2*2, which is 2³, which is 8.

This and that

There are 52 cards in a deck divided into four suits (i.e., hearts, diamonds, clubs, spades) of 13 cards each. What is the chance of me pulling out the ace of spades? It’s 1/52, of course.

What’s the chance of me pulling out the ace of spades and then any other spade? The chance of a pulling out the ace of spades is 1/52. The chance of then pulling out a second spade is 12/51, since there are only 12 spades remaining in the deck, and the deck now has 51 cards. The pulling of both cards are independent events. The combination of the two independent events is thus the multiple of the two: 1/52 x 12/51 = 0.0045, which is less than one half of one percent chance.

Probability theory writes this so, where P(A) is the probability of event A, P(B) is the probability of event B, and P(A and B) is the probability of both independent events happening:

P(A and B) = P(A) x P(B)

This and that… but this is one of those…

If the two events are not independent, meaning that one is not mutually exclusive from the other, the math above is slightly different. So, for example, the probability of drawing a card that is both a heart and a red king is not simply the multiple of the one event times the other. Because the red king is a heart, and the two are not mutually exclusive.

The probability of both dependent events happening is equal to the probability of one event, multiplied by the probability of the other, given the first. Or:

P(A and B) = P(A) x P(B|A)

So in this case, the probability of both dependent events happening is the probability of drawing a heart multiplied by the probability of it being a red king, given that we already know it’s a heart.

The probability of randomly drawing a heart, P(A), is 13/52. The probability of any given heart being a red king, P(B|A), is 1/13, since there is only one king of hearts out of all the 13 hearts. Thus, the probability of both dependent events is 13/52 x 1/13. Which is ~0.02, or a 2% chance.

This or that

The probability of doing this or that outcome, but not both, is the probability of doing this added to the probability of doing that. This applies to independent events. Represented in probability terminology thus:

P(A or B) = P(A) + P(B).

So the probability of pulling the king of hearts or any club is 1/52 + 13/52 = 14/52 = 0.27, where 1/52 is the probability of drawing the king of hearts, and 13/52 is the probability of drawing any club. This addition rule holds true for the probability of one outcome or another(s).

This or that… but this is one of those…

If the two events are not independent, meaning that one event is actually included in the outcome of another event, and the two are not mutually exclusive, then you get a slight variation on the math above. For example, the probability of drawing the king of hearts or any heart is different than the probability of drawing the king of hearts or any club. This is because the king of hearts is actually a heart. The probability for either of these dependent events is:

P(A or B) = P(A) + P(B) - P(A and B)

We know that the probability of drawing the king of hearts, P(A), is 1/52. We also know that the probability of drawing any heart, P(B), is 13/52, since there are 13 hearts in the deck. The probability of drawing a card that is both the king of hearts and a heart, P(A and B) where A and B are dependent, non-exclusive events (see above), is 1/52.

So the probability of either event, P(A or B), is 1/52 + 13/52 – 1/52, which equals 13/52.